地图api代码

Question

我错了，这是什么问题？我是怎么得到echo getAddress的?我是时候试着解决这个问题了？

function getAddress($la, $lo){
  $ur= "http://maps.googleapis.com/maps/api/place/nearbysearch/json?location=".$la.",".$lo."&radius=10&sensor=true&key=AIzaSyA0JD_Z2Uo2AfnDTejQFWHAXOIaRRpjF8c"
   $json = file_get_contents($ur);

   $data = json_decode($json);
   $status = $data->status;
   $name= '';
     if($status == "OK"){
      $addr = $data->results[0]->formatted_address;
     }
   return $addr;
  } 
          echo getAddress("6.154841","80.700845");

Answer 1

你忘了一个分号;在$ur的末尾

$ur= "http://maps.googleapis.com/maps/api/place/nearbysearch/json? location=".$la.",".$lo."&radius=10&sensor=true&key=AIzaSyA0JD_Z2Uo2AfnDTejQFWHAXOIaRRpjF8c";